Question: $f(x)=-(x-3)^2+25$ 1) What are the zeros of the function? Write the smaller $x$ first, and the larger $x$ second. $\text{smaller }x=$
Explanation: $\begin{aligned} -(x-3)^2+25&=0 \\\\ (x-3)^2&=25 \\\\ \sqrt{(x-3)^2}&=\sqrt{25} \\\\ x-3&=\pm 5 \\\\ x&=\pm5+3 \\\\ x={-2}&\text{ or }x={8} \end{aligned}$ $f(x)$ is given in vertex form: $f(x)=-(x-{3})^2+{25}$ So the vertex of the parabola is at $({3},{25})$. In conclusion, $\begin{aligned} \text{smaller }x&=-2 \\\\ \text{larger }x&=8 \end{aligned}$ The vertex of the parabola is at $(3,25)$